Respuesta :
Answer:
option d)378
Step-by-step explanation:
Given that a researcher at a major hospital wishes to estimate the proportion of the adult population of the United States that has high blood pressure.
Margin of error should be at most 6% = 0.06
Let us assume p =0.5 as when p =0.5 we get maximum std deviation so this method will give the minimum value for n the sample size easily.
We have std error = [tex]\sqrt{\frac{pq}{n} } =\frac{0.5}{\sqrt{n} }[/tex]
For 98%confident interval Z critical score = 2.33
Hence we have margin of error = [tex]2.33(\frac{0.5}{\sqrt{n} } <0.06\\n>377[/tex]
Hence answer is option d)378
The size of the sample needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 6% is; 376
What is the size of the sample?
We are told that Margin of error should be at most 6% = 0.06
Formula for margin of error is;
M = z√(p(1 - p)/n)
we are given the confidence level to be 98% and the z-score at this confidence level is 2.326
Since no standard deviation then we assume it is maximum and as such assume p =0.5 which will give us the minimum sample required.
Thus;
0.06 = 2.326√(0.5(1 - 0.5)/n)
(0.06/2.326)² = (0.5²/n)
solving for n gives approximately n = 376
Thus, the size of the sample required is 376
Read more about sample size at; https://brainly.com/question/14470673
